Download Trigonometrical functions

Mathematics 102 — Fall 1999
Trigonometry and calculus
Trigonometrical functions
In calculus, angles are always measured in radians. I recall that the measure of an angle in radians is the
ratio of an arc of a circle covered by the angle to its radius. Thus, for example, what you were taught in
grammar school to be a 360◦ angle is now 2π radians.
Exercise 1. How many radians is 180◦? 90◦ ?
For any number x, cos(x) and sin(x) are thus defined to be the horizontal and vertical coordinates of the
point on the unit circle at a distance x measured along the arc of the circle from the point (1, 0).
1
sin x x
x
cos x
By Pythagoras’ Theorem
cos2 x + sin2 x = 1
always. Two more formulas taht remain valid are the rule for the cosine and sum of angles:
cos(u + v) = cos u cos v − sin u sin v
sin(u + v) = sin u cos v + cos u sin v
Exercise 2. Find a formula for tan(u + v).
Exercise 3. Find a simple formula for 1 + tan2 (x).
Exercise 4. What is cos(π/2)? cos(π/4)? sin(π/4)? sin(−π/2)?
√
√
Exercise 5. An angle a has cos(a) = −1/ 2 and sin(a) = −1/ 2. What is a? Are there any other
possibilities?
Exercise 6. What is sin(π/6) (exactly)? Why? cos(π/6)? cos(π/3)? sin(π/3)?
Exercise 7. Describe all angles such that cos(a) = −1/2.
Their derivatives
2
Their derivatives
In order to calculate the derivative of sin(x) we have to apply the original idea of taking a limit of secant
slopes. We will first find sin0 (0) and cos0 (0). The second we can just read off from the graph.
Here are the graphs of both functions.
y = cos x
y = sin x
We can see immediately from the graph, if we did not know it already, that cos(0) = 1 is a maximum value
of cos(x), so that cos0 (0) has to be 0. The value of sin0 (0) is more difficult. Basically, the idea is that if an
angle is very small, then the arc length along that angle is just about the same as the vertical coordinate of
the corresponding point on the circle. In other words, as a first order approximation we have
sin(∆x) ∼ ∆x
which means that the limiting ratio of secant slope
sin(∆x)
sin(∆x) − sin(0)
=
∆x
∆x
is 1. So we have
cos(∆x) − cos(0)
∆x→0
∆x
=0
cos0 (0) = lim
sin(∆x) − sin(0)
∆x
sin(∆x)
= lim
∆x→0
∆x
=1
sin0 (0) = lim
∆x→0
Simply periodic functions
3
But now for any angle x at all we have
sin(x + ∆x) − sin(x)
∆x
sin(x) cos(∆x) + cos(x) sin(∆x) − sin(x)
= lim
∆x→0
∆x
cos(∆x) − 1
sin(∆x)
= sin(x) lim
+ cos(x) lim
∆x→0
∆x→0
∆x
∆x
sin0 (x) = lim
∆x→0
= sin(x) cos0 (0) + cos(x) sin0 (0)
= cos(x) .
The two limit expressions are the derivatives at 0 of sin(x) and cos(x). The derivative of cos(x) at 0 is 0
since cos(0) = 1 is the maximum possible value. And for ∆x very small we can see that sin(∆x) is just
about the same as ∆x, which means that the derivative of sin(x) at x = 0 is 1. Thus
sin0 (x) = cos(x) .
And since cos(x) = sin(π/2 − x), the derivative of cos(x) is
cos0 (x) = − sin0 (π/2 − x) = − cos(π/2 − x) = − sin(x) .
With this convention, the derivative of sin(x) turns out to be cos(x) and that of cos(x) is − sin(x).
Simply periodic functions
The functions cos x and sin x repeat every interval of 2π (because angles do). They rae said to be periodic.
In fact they are called simply periodic, because they are among the simplest of periodic functions, even in a
technical sense which I can’t explain here. The other simply periodic functions are the ones obtained from
these by scaling and shifting either vertically or horizonatlly. It turns out that all of them are of the form
C cos(ωx − α) for various constants C > 0, ω > 0, and α. This function oscillates between C and −C, and
for this reason C is called its amplitude. The constant ω is called its frequency; the period of the function
is 2π/ω.
What is not at all obvious is this: any combination A cos ωx + B sin ωx (two terms of the same frequency)
can be expressed as a simply periodic function of this form. It is just a matter of finding C and α. I will not
explain this completely here, but just show how the calculation works. We want to write
A cos ωx + B sin ωx = C cos(ωx − α) l.
To do this we apply the cosine sum formula on the righ to get
C cos(ωx) cos(α) + C cos(ωx) sin(α) = A cos ωx + B sin ωx,
which after equating corresponding coefficients leads to equations
A = C cos(α)
B = C sin(α)
2
A + B2 = C 2
p
C = A2 + b2
cos(α) = A/C
sin(α) = B/C
Functional inverses
4
This gives a direct formula for C, and exactly enough information to find the angle α.
Example. Look at
cos x + sin x .
√
2
√
cos(α) = 1/ 2
√
sin(α) = 1/ 2
α = π/4
We have
C=
so
cos x + sin x =
√
2 cos(x − π/4) .
Functional inverses
If f (x) is any function then another function F (x) is called its functional inverse if f (F (x)) = x. In other
words, if we first apply f and then F we get back where we started from. Equivalently, this means that for
any number y we get F (y) by solving for y in the equation y = f (x).
For example, if f (x) = ex then we solve
y = ex ,
x = log(y)
so that log is the functional inverse of the exponential function. We disciussed this before, from a slightly
different point of view—we saw that the graph of y = log(x) is obtained from the graph of y = ex by
swapping x and y.
Similarly,
y = x2 ,
x=
√
y
so the square root is the functional inverse of the square. Let’s look at this example a bit more. Here is the
graph of y = x2 :
And here is what we get when we swap x and y:
The functional inverse of tan(x)
5
y=
√
x
√
There are a few problems. The first is that not every number has a square root—the domain of y = x, the
set of numbers for which it is defined, is the range of non-negative numbers. A second problem is that even
for x > 0 there isn’t a unique square root of x, but two So we must pick one—by convention the non-negative
2
we must add a bit to the description. For x ≥ 0
one. So in order
√ to define the functional inverse of y = x √
the number x is the unique non-negative one such that ( x)2 = x.
In any event, we are interested in finding a formula for the derivative of a functional inverse. By the chain
rule, we have
f (F (x)) = x
f 0 (F (x))F 0 (x) = 1
F 0 (x) =
1
f 0 (F (x))
For example, the functional inverse of ex is log(x), so
1
elog(x)
1
=
x
log0 (x) =
since the derivative of ex is ex .
The functional inverse of tan(x)
Recall that
tan(x) =
sin x
.
cos x
We can calculate the derivative by the quotient rule
tan0 (x) =
1
.
cos2 x
Exercise 8. Sketch the graph of y = tan(x) for x = −π/2 to π/2.
We define arctan(x) to be the unique number y such that y lies between −π/2 and π/2, and tan(y) = x.
Exercise 9. Sketch the graph of y = tan(x) for x = −10 to 10.
Exercise 10. Sketch the graph of y = arctan(x) for x = −10 to 10.
The functional inverse of tan(x)
6
As for its derivative, if F (x) = arctan(x) then
F 0 (x) =
1
cos2 (arctan(x))
√
1+
x2
since the derivative of tan(x) is 1/ cos2 (x). This formula can be simplified, since the following picture shows
that
1
.
cos(arctan(x)) = √
1 + x2
x
1
Exercise 11. What is arctan(1)? arctan(0)? arctan(−1)?
Exercise 12. What happens to arctan(x) as x → ∞ (i.e. x gets larger and larger positive)?
For any x with −1 ≤ x ≤ 1 arcsin(x) is defined to be the unique number y such that π/2 ≤ y ≤ π/2 and
sin(y)
√= x. It is a functional inverse of sin(x). A similar calculation will show that the derivative of arcsin(x)
is 1/ 1 − x2 .
y = arcsin(x)
Exercise 13. Define arccos(x) and find its derivative.
Exercise 14. What is arcsin0 (1)?
The table of derivatives so far
7
The table of derivatives so far
Recall that calculating derivatives is a recipe in two parts. The first part is memorizng the derivatives of a
handful of functions that are in some sense basic. The second involves applying a set of four or five rules
(sum rule, product rule, chain rule, quotient rule, and now in addition the functional inverse rule, even
though it is a consequence of the chain rule) to the functions in this table. In this chapter we have in effect
added some functions to the table. Here is what is it is now:
f (x)
constant
xr
ex
cos(x)
sin(x)
f 0 (x)
0
rxr−1
ex
− sin(x)
cos(x)
In addition, there are a few that you might want to memorize even though they can be derived from the rule
for the derivatives of functional inverses:
log(x)
arctan(x)
arcsin(x)
1/x
2
1/(1
+
√ x )
1/ 1 − x2